## derivative of cotx

### To find the derivative of cot(x), we need to use the quotient rule, which states that if we have the function f(x) = g(x) / h(x), then its derivative f'(x) is given by:

f'(x) = (g'(x) * h(x) – g(x) * h'(x)) / [h(x)]^2

To find the derivative of cot(x), we need to use the quotient rule, which states that if we have the function f(x) = g(x) / h(x), then its derivative f'(x) is given by:

f'(x) = (g'(x) * h(x) – g(x) * h'(x)) / [h(x)]^2.

Let’s apply this rule to find the derivative of cot(x):

First, let’s define the cotangent function in terms of sine (sin(x)) and cosine (cos(x)). The cotangent function is defined as cot(x) = cos(x) / sin(x).

Using the quotient rule, we differentiate the numerator (cos(x)) and the denominator (sin(x)) separately.

For the numerator:

The derivative of cos(x) is -sin(x). This is because the derivative of cos(x) is obtained by taking the derivative of the inside function (x) which is -sin(x), multiplied by the derivative of the inside function, which is 1.

For the denominator:

The derivative of sin(x) is cos(x). This is because the derivative of sin(x) is obtained by taking the derivative of the inside function (x) which is cos(x), multiplied by the derivative of the inside function, which is 1.

Now, we can use the derivative rule and substitute the derivatives we found into the quotient rule formula:

f'(x) = (-sin(x) * sin(x) – cos(x) * cos(x)) / [sin(x)]^2

Simplifying further:

f'(x) = -(sin^2(x) + cos^2(x)) / [sin^2(x)]

Using the trigonometric identity sin^2(x) + cos^2(x) = 1:

f'(x) = -1 / [sin^2(x)]

Alternatively, you can rewrite the result using the identity cot(x) = 1 / tan(x):

f'(x) = -1 / [sin^2(x)] = -1 / [(1/cos^2(x))] = -cos^2(x)

Therefore, the derivative of cot(x) is -cos^2(x).

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